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Homework Help: Arguments of Roots of Complex Numbers

  1. May 15, 2010 #1
    1. The problem statement, all variables and given/known data
    i) Solve the equation [itex] z^3 = \mathbf{i}[/itex].
    (ii) Hence find the possible values for the argument of a complex number w which is such that [itex] w^3 = \mathbf{i}(w*)^3[/itex].

    I'm stuck on part ii.

    2. Relevant equations


    3. The attempt at a solution

    The answer to the equation in part i: e^(1/6 pi i) , e^(5/6pi i ), e^(-1/2pi i )


    In the back of the book it has 6 answers, I can only get three of them. If I draw it out on the argand diagram the 3 answers I am missing would be the arguments if you carried the line on through the axis. For example I have arguments 5/12pi , pi/12, -pi/4 and am missing 3/4pi, -11/12pi and -7/12pi.

    My method was to cube root both sides:
    e^theta i = Z e^ -theta i
    Where Z is one of the solutions to part i, by changing Z I can get 3 solutions but am missing the other three.

    After seeing in the book there were six solutions I did manage to get them by letting i = e^pi/2 i and setting it equal to e^6theta i, then I just kept adding 2pi to the argument of i until I got all 6 arguments and ended up back at the beginning. However if I didn't have the answers I wouldn't have known to look for the extra three, and this method didn't involve part i) of the question (which I think it is meant to because of the 'hence').

    I'd be interested to see how you all approach the problem, and an explanation of what is actually happening.


    Thanks
     
  2. jcsd
  3. May 15, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi I like number! Welcome to PF! :smile:

    (have a pi: π and try using the X2 tag just above the Reply box :wink:)

    Hint: write it (w/w*)3 = i :wink:
     
  4. May 15, 2010 #3
    Re: Welcome to PF!

    Thanks for the reply.
    Doing that I end up with

    e2itheta=3rdrt i

    From part i) of the question I have all 3 values for the 3rdrt i , so i set e2itheta= to each of them one at a time , then find theta but this only gives 3 values.
    If I was to plot the values I get on an argand diagram mine would be in the first and fourth quadrant, and I'm missing the 3 values from the second and third quadrants.

    I hope I'm making sense, and thanks once again for the help.

    I like number
     
  5. May 15, 2010 #4

    tiny-tim

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    ok, so you have three values for e2iθ, which is (e)2,

    so e is the square-root of that, which has two values, ± of each other. :wink:
     
  6. May 16, 2010 #5
    :blushing: *doh*!

    e= ± eπ/12i
    And from this I get the value I had which was π/12 if I take the positive root, if I take the negative root then the argument will be π12 + π = 13π/12 which gives the argument -11π/12 which was one of the missing ones :D If i do this for the others I'll get them :)
    Perfect, thank you tiny-tim!

    I like number
     
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